Classical statistic: how do we derive $\langle E\rangle=kT$ (where $k$ is the Boltzmann constant)?

Question

I am following a propaedeutic course in quantum mechanics, and we did some basics of statistical mechanics deriving the Boltzmann equation for energy: $$n_i=\frac{N}{Z}g_i e^\frac{-E_i}{kT}$$ where $n_i$ is the number of particles with $E_i$, $Z$ is the partition function, and $g_i$ is the multiplicity. From that equation, we derived that the average energy is: $$\bar E=kT^2\frac{d}{dT}(\log Z)$$ Some lessons later, we are now doing Rayleigh-Jeans model for the black body, and we said that the electrons on the walls of the black body are modelled as 1-dimensional harmonic oscillators, so, for the equipartition theorem, the average kinetic energy is $\frac{1}{1}kT$, the average potential energy is $\frac{1}{1}kT$, and the total energy is $kT$. After that, our professor told us this can be derived also from statistical mechanic, because $$p(E)=\frac{e^\frac{-E_i}{kT}}{kT}$$ where kT is the normalization factor, and, by definition of average: $$\bar E=\int_{0}^\infty Ep(E)dE=kT$$ But now I am confused: where did the factor $\frac{N}{Z}g_i$ go? And therefore, where did the factor $\frac{d}{dT}(\log Z)$ go when calculating $\langle E\rangle$?

Answer 1

Probably, part of the explanation passed unnoticed, or it was missing. The formula $$ p(E)=\frac{e^{-\frac{E}{kT}}}{kT} \tag{1} $$ has not the same validity as the general probability of finding the system in a state of energy $E$ in a canonical ensemble, which can be obtained from your formula for $n_i$ dividing by $N$.

However, once one has justified the use of the partition function for a set of non-interacting one-dimensional harmonic oscillators$^*$, formula ($1$) is the correct energy distribution for the $1D$ harmonic oscillator, the energy density of states is a constant (it does not depend on $E$). Therefore, if $g_i=constant$, the probability is proportional to $e^{-\frac{E}{kT}}$, and the normalization factor of the density distribution in energy is $\frac{1}{kT}$. If we evaluate the partition function of a classical harmonic oscillator in $1D$, we find $$ Z=\frac{kT}{\hbar \omega}, \tag{2} $$ differing from the previous normalization by an energy factor $\hbar \omega$, necessary for dimensional reasons (the probability density of states in the phase space is dimensionless, while the probability density of energy has the dimension of inverse energy). In any case, a partition function depending linearly on $T$ is all we need to get the average energy equal to $kT$.

Summarizing, equation $(1)$ is valid only for the $1D$ classical harmonic oscillator. However, the final result $\left< E \right>= kT$, is a special case of the equipartition theorem, stating that the equilibrium average of each quadratic degree of freedom in the Hamiltonian is equal to $\frac{kT}{2}$.

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$^*$ I'm afraid I disagree with the justification for using a set of harmonic oscillators the way you reported (electrons on the walls of the black body are modeled as 1-dimensional harmonic oscillators). The main reasons are the following:

1. In the case of blackbody radiation, the system of interest is the radiation field in the cavity, not the electrons;
2. It is the electromagnetic field, expressed in terms of normal modes, that is described exactly as a set of non-interacting harmonic oscillators. Any electronic system could be only approximately treated that way.
3. The so-called ultraviolet catastrophe is due to the infinite number of normal modes required to describe the field in the cavity. The number of electrons, even if macroscopical, is finite.